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EXPLANATION OF ASTATINE IONIZATIONS
By Prof.L. Kaliambos (Natural Philosopher in New Energy) June 23 , 2015 ' INTRODUCTION' Astatine is a chemical element with symbol At and atomic number 85. Despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of atoms with many electrons related to the chemical properties. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity.(EXPERIMENTS REJECTING EINSTEIN). It is of interest to note that the discovery of the electron spin by Uhlanbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin.( FASTER THAN LIGHT). So it was my published paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of astatine with the following electron configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d10 6s2 6px2 6py2 6pz1 According to the “Ionization Energies (eV) of Atoms and Ions” the ionization energies of astatine (from E1 to E5 ) are the following: E1 = 9.3 , E2 = 20 , E3 = 29 , E4 = 41 , and E5 = 51 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my new paper of 2008. ' ' EXPLANATION OF E1 = 9.3 eV = -E(6px2) + E(6px1) The electron charges (-80e) of the 80 electrons of the following configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f145d106s2) screen the nuclear charge (+85e) and for a perfect screening the electrons of 6p5 would provide an effective Zeff = ζ = 5. But the five electrons of 6p repel the 6s2 and the 5d10 and lead to the deformation of spherical electron clouds. Thus ζ > 5. Here the E(6px2) represents the binding energy of 6px2, while the E(6px1) represents the binding energy of 6px1, which appears after the first ionization of 6px2 . Note that the 6px2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6px2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6px1 consists of one electron, we apply the Bohr formula as E(6px1) = (-13.6057)ζ2/n2 Therefore E1 = 9.3 eV = -E(6px2) + E(6px1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 330.7 = 0 and solving for ζ we get ζ = 5.59 > 5. Of course the two electrons of opposite spin (6px2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that two electrons with opposite spin exert a mutual Coulomb repulsion. Under such fallacious ideas I published my paper of 2008 . EXPLANATION OF E2 = 20 eV = -E(6py2) + E(6py1) As in the case of E1 the electron charges (-80e) of the 80 electrons of the following configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f 145d106s2) screen the nuclear charge (+85e) and for a perfect screening after the first ionization the four electrons of 6p5 would provide the same effective ζ = 5. But the four electrons of 6p repel the 6s2 and the 5d10 and lead to the deformation of spherical electron clouds. Thus ζ > 5. Here the E(6py2) represents the binding energy of 6py2, while the E(6py1) represents the binding energy of 6py1, which appears after the first ionization of 6py2 . Note that the 6py2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(6py2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 6py1 consists of one electron, we apply the Bohr formula as E(6py1) = (-13.6057)ζ2/n2 Therefore E2 = 20 eV = -E(6py2) + E(6py1) = - (16.95)ζ + 4.1) / n2 Then using n = 6 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 715.9 = 0 and solving for ζ we get ζ = 7.9 > 5. Here the ζ = 7.9 > 5.59 > 5 means that after the ionization the four electrons of 6p5 break the symmetry and lead to a greater deformation of electron clouds. EXPLANATION OF E3 = 29 eV = -E(6px1) As in the cases of E1 and E2 the electron charges (-80e) of the 80 electrons of the following electron configuration (1s22s22p63s23p63d104s24p64d105s25p6 4f145d106s2) screen the nuclear charge (+85e) and for a perfect screening after the two ionizations the three electrons of 6p5 would provide the same effective ζ = 5. But after the two ionizations the three electros of 6p with parallel spin repel the 6s2 and the 5d10 and lead to the deformation of electron clouds. Thus ζ > 5. Here the E(6px1) represents the binding energy of 6px1, given by applying the Bohr formula as E3 = 29 eV = E(6px1) = (-13.6057)ζ2/n2 Then using n = 6 we get ζ = 8.76 > 5. Here the ζ = 8.76 > 7.9 > 5.59 > 5 means that after the ionizations the three electrons of 6p5 break the symmetry and lead to a greater deformation of electron clouds. ' ' EXPLANATION OF E4 = 41 eV = -E(6py1) Here the E(6py1) represents the binding energy of 6py1 given by applying the Bohr formula as E4 = 41 eV = -E(6py1) = -(-13.6057)ζ2/n2 Then using n = 6 the above equation can be written as E4 = 41 eV = -E(6py1) = -(-13.6057)ζ2/ 62 and solving for ζ we get ζ = 10.4 > 5 . Here the ζ = 10.4 > 8.76 > 7.9 > 5.59 > 5 means that after the ionizations the two electrons of 6p5 break more the symmetry and lead to a greater deformation of electron clouds. EXPLANATION OF E5 = 51 eV = -E(6pz1) Here the E(6pz1) represents the binding energy of 6pz1 given by applying the Bohr formula as E3 = 51 eV = -E(6pz1) = -(-13.6057)ζ2/n2 Then using n = 6 the above equation can be written as E5 = 51 eV = -E(6pz1) = -(-13.6057)ζ2/ 62 and solving for ζ we get ζ = 11.6 > 5 . Here the ζ = 11.6 > 10.4 > 8.76 > 7.9 > 5.59 > 5 means that after the ionizations the one electron of 6p5 breaks more the symmetry and leads to a greater deformation of electron clouds. Category:Fundamental physics concepts